**Interest (I):**It is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed

**Compound Interest (CI)**

CI = P(1+R/100)^{T} – P

A = P(1+R/100)^{T}

Where,

P= Principle or sum {It is the money borrowed or lent out for a time period T}

R=Rate at which interest is charged on P per annum {Thus, 6 p.c. means that Rs.6 is the interest on Rs.100 in one year.}

T=time

A = Amount {The Addition of Interest and Principle i.e. A = (I + P)}

Compound Interest is based on the principal amount and the accumulated interest.

**Compound Interest questions can be designed in many ways. Lets discuss all one by one:**

**Case 1: When compound Interest is calculated on yearly basis.**

Compound Interest is calculated at the end of one year duration and

now for next year CI+P = next year principle.

Let principle = P, time = T years and rate = R% per annum and

Let A be the total amount at the end of T years, then

A = P (1+R/100)^{T}

CI = P(1+R/100)^{T} – P {Important to note}

**Example 1.** An amount of Rs.8000 is invested in a FD scheme for 2 years at compound interest rate 5% per annum. How much interest will be incurred on maturity of the FD?

**Explanation:**

I = 8000(1+5100)^{2} – 8000

= 8000(105100)^{2} – 8000

= 8000 [105100)^{2} -1]
= 8000 [1.103-1]
= 8000*0.103

= 824

Case 2: When compound interest is calculated bi-yearly (6 months).

Compound Interest is calculated at the end of 6 months duration from investment and now for 6 months CI+P = next year principle.

If the annual rate is R% per annum and is to be calculated for T years, then in this case,

rate = (R/2) half-yearly and,

time = (2T) 2 half-years).

CI = P [1+(R/2)/100]^{T/2} – P

**Example 2.** An amount of Rs.15,000 is invested at 10% per annum for one year. If the interest is compounded half-yearly, then what amount will be received at the end of the year?

**Explanation:**

A = 15000 [1+(10/2)/100]^{2}

= 15000 [105/100]^{2}

= 15000 X 1.103 (Click to see short trick for finding square quickly)

= 16545

**Case 3: When compound interest is reckoned quarterly.**

In this case, rate = (R/4)% quarterly and,

time = (4T) quarter years.

A= P[1+(R/4)/100]^{4T}

**Example 3.** Find the compound interest on Rs. 15,000 for 9 months at 16% per annum compounded quarterly.

**Explanation:**

CI = 15000[1+(16/4)/100]^{4X(9/12)} -15000

= 15000[104/100]^{3} – 15000

= 15000 [1.120-1] (click here to learn quick cubes)

= 15000 X 0.120

= 1800

** Case 4: When rates are different for different years, say for 1st, 2nd, 3rd year R1, R2, R3 respectively. Then**

A = P (1+R1/100) (1+R2/100) (1+R3/100)

**Example 4.** An amount of 200000 is deposited in bank under a scheme which provides 10% rate 1st years, 20% rate 2nd year and 5% rate 3rd year. After 3rd year what amount will be returned?

**Explanation: **

A = 200000 (1+10/100)(1+20/100)(1+5/100)

A= 200000(110/100)(120/100)(105/100)

A= 277200

**Some Tricks to Solve easily **

*Trick 1:* If a sum of money becomes “n” times in “T years” at simple interest, then the rate of interest per annum can be given by:

R=(n-1)*100/T

*Explanation:*

A=nP

SI=A-P=nP-P=(n-1)P

SI=PRT/100

(n-1)P=PRT/100

R=(n-1)*100/T

*Trick 2:* If an amount P_{1} is lent out at simple interest of R1% per annum and another amount P_{2} at simple interest rate of R_{2}% per annum, then the rate of interest for the whole sum can be given by:

R=(P_{1}.R_{1} + P_{2}.R_{2})/(P_{1}+P_{2})

*Explanation:*

SI_{1}=P_{1}.R_{1}.T/100

SI_{2}=P_{2}.R_{2}.T/100

If P_{1} and P_{2} are lent at rate R so that

SI_{1}+SI_{2}= SI then,

SI=(P_{1}+P_{2})R.T/100

SI_{1}+SI_{2}= (P_{1}+P_{2})R.T/100

P_{1}.R_{1}.T/100 + P_{2}.R_{2}.T/100 = (P_{1}+P_{2})R.T/100

R=(P_{1}.R_{1} + P_{2}.R_{2})/(P_{1}+P_{2})

*Trick 3:* In what time will the simple interest becomes “n times” of the principal P at “R %” per annum:

T =n x 100/R

*Explanation:*

nP=PRT/100

T=n*100/R

*Trick 4:* If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R_{1}, R_{2}, … , R_{n} respectively and time periods are T_{1}, T_{2}, … , T_{n} respectively, then the ratio in which the sum will be divided in n parts can be given by

1/R_{1} . T_{1} : 1/R_{2} . T_{2} : ……….. : 1/R_{n} . T_{n}

*Explanation:*

P=P_{1}+P_{2}+P_{3}………….+P_{n}

SI=P_{1}.R_{1}.T_{1}/100

P_{1}=SI.100/R_{1}.T_{1}, P_{2}=SI.100/R_{2}.T_{2}, P_{3}=SI.100/R_{3}.T_{3}………….., P_{n}=SI.100/R_{n}.T_{n}

P_{1} : P_{2} : P_{3} : ……………………. : P_{n}

SI.100/R_{1} . T_{1} : SI.100/R_{2} . T_{2} : SI.100/R_{3} . T_{3} : ………….. : SI.100/R_{n} . T_{n}

**Difference between CI and SI:**

Shortcut Trick:

For two years:

CI-SI = P(R/100)^{2}

For three years:

CI-SI = 3P(R/100)^{2} + P (R/100)^{3}

**Explanation: **

**For two years: **

CI-SI =[P(1+R/100)^{2} – P ]- 2PR/100

= [P(1+2R/100+R^{2}/(100 . 100)) – P] -2PR/100

= [P + 2PR/100 + PR^{2}/(100 . 100)-P] – 2PR/100

= [2PR/100 + PR^{2}/(100 . 100)] – 2PR/100

= P(R/100)^{2}

**For three years:**

Similarly we can calculate the difference for three years:

CI-SI = [P(1+R/100)^{3} – P ]- 3PR/100

= [P((1+R/100) (1+2R/100+R^{2}/(100 . 100)) – P] -3PR/100

= [P+2PR/100+PR^{2}/(100 . 100)+PR/100+2PR^{2}/100^{2}+PR^{3}/100^{3}-P] – 3PR/100

= [3PR/100+3PR^{2}/100^{2}+PR^{3}/100^{3}] – 3PR/100

= 3P(R/100)^{2} + P (R/100)^{3}

## Comments 3

Then what for n years

Author

Generally in competitive exams upto 3 years is asked. But we can devise the same for n also in the similar fashion.

Then what will be the difference of ci and si for n years